Velocity-time graphs

If an object moves along a straight line, its motion can be represented by a velocity-time graph. The gradient of the line is equal to the acceleration of the object.

A velocity/time graph. Graph with four distinct sections. All lines are straight.

The table shows what each section of the graph represents:

Section of graphGradientVelocityAcceleration
Dzerostationary (at rest)zero


Describe the motion of the vehicle in the graph at the three stages of its journey.

Between 0 s and 4 s the vehicle is accelerating.

acceleration = \frac{change~in~velocity}{time~taken}

This means that: 8 \div 2 = 4 m/s^2

Between 4 s and 7 s the vehicle has a constant velocity of 8 m/s.

Between 7 s and 10 s, the vehicle is decelerating so: -8 \div 3 = -2.67~m/s^2

Calculating displacement - Higher

The displacement of an object can be calculated from the area under a velocity-time graph.

The area under the graph can be calculated by:

  • using geometry (if the lines are straight)
  • counting the squares beneath the line (particularly if the lines are curved)


Calculate the total displacement of the object whose motion is represented by the velocity-time graph below.

The y axis shows velocity in metres per second and the x axis time in seconds.  The object increases its velocity from 0 metres per second to 8 metres per second in 4 seconds.

The displacement can be found by calculating the total area of the shaded sections below the line.

Find the area of the triangle:

\frac{1}{2} \times base \times height

\frac{1}{2} \times 4 \times 8 = 16~m^{2}

Find the area of the rectangle:

base × height

(10 - 4) × 8 = 48 m2

Add the areas together to find the total displacement:

(16 + 48) = 64 m

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